Question

derivation of three questions of motion,graphically

please tell me fast and make graph​

Answer 1

Answer:

I think you wrote the equation but the question was written by mistake

An object is in motion with initial velocity u attains a final velocity v in time t due to acceleration a, with displacement s. Let us try to derive these equations by graphical method.

Graph shows the change in velocity with time for an uniformly accelerated object. The object starts from the point D in the graph with velocity u. Its velocity keeps increasing and after time t it reaches the point B on the graph. The initial velocity of the object = u = OD = EA The final velocity of the object = v = OC = EB Time = t = OE = DA Also from the graph we know that, AB = DC

For First equation of motion By definition, acceleration = change in velocity / time = (final velocity – initial velocity)/time = (OC – OD) / OE = DC / OE a = DC/t DC = AB = at From the graph EB = EA + AB v = u + at …….(1) This is first equation of motion. For Second equation of motion From the graph the distance covered by the object during time t is given by the area of quadrangle DOEB s = area of the quad-rectangle DOEB s= area of the rectangle DOEA + area of the triangle DAB (AE × OE) + (1212 × AB × DA) s = ut + 1212at2 ….(2) This is second equation of motion. For Third equation of motion From the graph the distance covered by the object during time t is given by the area of thequadrangle DOEB. Here DOEB is a trapezium. Then s = area of trapezium DOEB = 1212× sum of length of parallel side × distance between parallel sides = 1212 × (OD + BE) × OE s = 1212 × (u + v) × t since a = (v – u) / t or t = (v – u)/a Therefore s = 1212 × (v + u) × (v – u)/a 2as = v2 – u2 v2 = u2 + 2 as ………..(3) This is third equation of motion...

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